How to calculate the following limit - I'm stuck

[chemic_23]

Homework Statement

what is the limit of

as x approaches 9?

Homework Equations

The Attempt at a Solution

I'm stuck please help...

 

[sara_87]

I would do it this way:
remember that:
a$$^{2}$$-b$$^{2}$$=(a+b)(a-b)
so
a-b=$$\frac{(a^2-b^2}{(a+b)}$$

so take the numerator and denominator of your problem and treat each as a-b
so:
$$\sqrt{x}-3$$ = $$\frac{x-9}{\sqrt{x}+3}$$ (1)

and
$$\sqrt{1+\sqrt{x}}-2$$ = $$\frac{\sqrt{x} - 3}{\sqrt{1+\sqrt{x}}+2}$$ (2)

so your limit is now (1)/(2): which gives:

$$\frac{x-9}{\sqrt{x}+3}\times\frac{\sqrt{1+\sqrt{x}}+2}{\sqrt{x}-3}$$

Is this clear?
Do you know how to continue from here?

 

[chemic_23]

continuation...

thus, the limit of

as x approaches 9 is 4. Is this correct?

 

[Dick]

Correct.